/sci/ - Science & Math

>>11489318
Because it determines an orientation on the top exterior algebra [math]\Lambda^3 T^*\mathbb{R}^3[/math] and hence a volume form [math]\operatorname{vol}\in \Omega^3(\mathbb{R}^3)[/math]. Whenever you do integration, we're orienting our spaces, surfaces and lines with respect to this volume form.
You can certainly also use the left hand rule, but it reverses the orientation. All it does is introducing a minus sign on our orientation and integrals.

>>11332984
Close but not quite.

>>11261071
I don't know where you get your last relation, but the momentum operator is typically [math]\hat{p} = -i\partial_x[/math] from holomorphic quantization. When you make this replacement (called the first quantization) in the relativistic relation [math]E^2 = m^2 + p^2[/math] ([math]c=1[/math] in natural units) you get the Klein-Gordon operator [math]\hat{H}^2 = \square^2 + m^2[/math], which is the [math]square[/math] of the Hamiltonian/total energy. The reason the Klein-Gordon equation is not sufficient is because [math]\hat{H} = \sqrt{\square^2+m^2}[/math] is not a compact operator, and hence does not inherit the well-posedness and hyperbolicity of the Klein-Gordon operator. It also leads to paradoxes like the Klein paradox and Eckstein's no-go theorem. This means first quantization must be performed on spin-manifolds from which we obtain the Dirac operator [math]\not\partial - im[/math]. Quantization is much more subtle in this case.